Solve 1/3(x+1)-1/4(x+4)geq5/6x-1

$$\frac { 1 } { 3 } ( x + 1 ) - \frac { 1 } { 4 } ( x + 4 ) \geq \frac { 5 } { 6 } x - 1$$

Solve for x

$$x\leq \frac{4}{9}$$

Solution Steps

Use the distributive property to multiply \frac{1}{3} by x+1.

$$\frac{1}{3}x+\frac{1}{3}-\frac{1}{4}\left(x+4\right)\geq \frac{5}{6}x-1$$

Use the distributive property to multiply -\frac{1}{4} by x+4.

$$\frac{1}{3}x+\frac{1}{3}-\frac{1}{4}x-\frac{1}{4}\times 4\geq \frac{5}{6}x-1$$

Cancel out 4 and 4.

$$\frac{1}{3}x+\frac{1}{3}-\frac{1}{4}x-1\geq \frac{5}{6}x-1$$

Combine \frac{1}{3}x and -\frac{1}{4}x to get \frac{1}{12}x.

$$\frac{1}{12}x+\frac{1}{3}-1\geq \frac{5}{6}x-1$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{1}{12}x+\frac{1}{3}-\frac{3}{3}\geq \frac{5}{6}x-1$$

Since \frac{1}{3} and \frac{3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{12}x+\frac{1-3}{3}\geq \frac{5}{6}x-1$$

Subtract 3 from 1 to get -2.

$$\frac{1}{12}x-\frac{2}{3}\geq \frac{5}{6}x-1$$

Subtract \frac{5}{6}x from both sides.

$$\frac{1}{12}x-\frac{2}{3}-\frac{5}{6}x\geq -1$$

Combine \frac{1}{12}x and -\frac{5}{6}x to get -\frac{3}{4}x.

$$-\frac{3}{4}x-\frac{2}{3}\geq -1$$

Add \frac{2}{3} to both sides.

$$-\frac{3}{4}x\geq -1+\frac{2}{3}$$

Convert -1 to fraction -\frac{3}{3}.

$$-\frac{3}{4}x\geq -\frac{3}{3}+\frac{2}{3}$$

Since -\frac{3}{3} and \frac{2}{3} have the same denominator, add them by adding their numerators.

$$-\frac{3}{4}x\geq \frac{-3+2}{3}$$

Add -3 and 2 to get -1.

$$-\frac{3}{4}x\geq -\frac{1}{3}$$

Multiply both sides by -\frac{4}{3}, the reciprocal of -\frac{3}{4}. Since -\frac{3}{4} is negative, the inequality direction is changed.

$$x\leq -\frac{1}{3}\left(-\frac{4}{3}\right)$$

Multiply -\frac{1}{3} times -\frac{4}{3} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-\left(-4\right)}{3\times 3}$$

Do the multiplications in the fraction \frac{-\left(-4\right)}{3\times 3}.

$$x\leq \frac{4}{9}$$